Coursera: analysis complex kind Week 3.1. Differentiability, continuity, analytic functions

Jul 17, 2016

Key definitions in this post.

Slide 2

Reviewed the definition of real derivative of f(x) at x_0 as limit

\underset{x \to {x_0}}{\lim} \, \frac{f(x) - f(x_0)}{x - x_0}

Slides 3 & 4

Graphical examples given as secant becomes tangent.

Slide 5

“Failing” Example when graph has a “point”, that is the gradient of the curve is different each side so gradient limit from either side of x0 is not the same, therefore derivative at x0 does not exist.

Slide 6

Complex derivative is essentially the same

\underset{z \to {z_0}}{\lim} \, \frac{f(z) - f(z_0)}{z - z_0}

… different conventions that represent derivative here ..

or, if h is a small complex number, and we write z as z_0 + h the limit can be written as

\underset{h \to 0}{\lim} \, \frac{f({z_0}+h) - f(z_0)}{h}

Since h is any complex number, z_0 can be “approached” from any  “direction”.

 

This definition of the complex derivative as a limit with same “form” as real derivative leads to (?) the same standard results as in real-valued calculus

e.g. d/dz(c f(z)) = c f ‘(z), c is a constant

  • addition rule applies
  • product rule applies
  • chain rule applies

 

as a result complicated functions, e.g. z^2 e^{sin z} can sometimes be split into simpler components for differentiation.

*Important* Slide 11 – non examples

Consider f(z) = Re(z)

If z = x + iy, so f(z) = Re(z) = x and let h = h_x + i h_y

\frac{{f({z}}+{h}) - f(z)}{h} = \frac{(x+h_x) - x}{h}= \frac{h_x}{h}= \frac{Re(h)}{h} (*)

So .. does this have a limit as h -> 0?

h-> 0 along real axis. Then h = h_x + i0 = h_x

Substituting into (*) we get h / h = 1

Whereas, h-> 0 along imaginary axis. Then h = 0 + ih_y = i h_y

Since Re(h) now equals zero, when we substitute this into (*) we get a quotient value of zero.

So depending on how h approaches zero (along real or imaginary axes) the difference quotient has two values, 1 and zero, so f is not differentiable at z = 0.

Further …

Consider h to be a sequence of values that spiral into towards the origin but alternating between real and imaginary values

<< sketch here would help >>

A suitable function (which we put in place of h) could be

h_n = \frac{i^n}{n}, where n is a positive integer, creating the sequence

\frac{i^1}{1},\frac{i^2}{2},\frac{i^3}{3},\frac{i^4}{4},\frac{i^5}{5}, ... = i, \frac{-1}{2},\frac{-i}{3},\frac{1}{4},\frac{i}{5}, ...

So that sequence of h values is spiralling towards zero

And the difference quotient (*) becomes \frac{Re \; h_n}{h_n} \; = \; \frac{Re \; i^n}{i^n}

Which has the values

1 when n is even:  \frac{ Re \; i^2}{i^2},  \frac{Re \; i^4}{i^4}, ... =  \frac{-1}{-1},  \frac{1}{1},, ...

0 when n is odd  \frac{ Re \; i^1}{i^1},\frac{Re \; i^3}{i^3},\frac{Re \; i^5}{i^5}, ... = 0, \frac{0}{-i},\frac{0}{i}, ...

 

so the sequence has no limit, and since z was arbitrary, function is not differentiable anywhere on C.

Slide 12 Another non-example f(z) = \overline{z}

Let f(z) = \overline{z}

So defining z and h as in slide 11

 \frac{f({z}+h) - f(z)}{h} = \frac{(\overline{z}+\overline{h}) - (\overline{z})}{h}= \frac{\overline{h}}{h}

If h \in R then \frac{\overline{h}}{h} = 1 -> 1 as h -> 0.

If h \in iR then \frac{\overline{h}}{h} = -1 -> -1 as h -> 0.

Thus \frac{\overline{h}}{h} does NOT have a limit as h -> 0, and since z is arbitrary, f (complement of z) is not differentiable anywhere in C.

Slide 13 (complex) Differentiability implies continuity

** Slide 14 Analytic functions **

Definition : A function is analytic in an open subset U ⊂ C, if it is complex differentiable at every point in U

Definition: A function which is analytic on all of C is entire.

Examples:

Polynomials analytic every in C therefore entire

Rational functions p(x)/q(x) are analytic when q(x) not equal zero.

Modulus function not analytic

Re(z), Im(z) not analytic

Another example, by first principles f(z) =( modulus(z))^2 is differentiable only at the single point z=0 therefore not analytic, however it is continuous everywhere in C

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2 Responses to “Coursera: analysis complex kind Week 3.1. Differentiability, continuity, analytic functions”


  1. […] f:U -> C is analytic  and if U such that then  f is conformal […]


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